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Can we prove the following is always an integer?

$$6b\sum_{k=1}^bk\left\{\frac{ka}{b}\right\}$$

where $\{x\}=x-\lfloor x\rfloor$ denotes the fractional part operator.

UPDATE:

Through the calculation of several identities and going through heavy casework, I have proven the above formula; the proof, however, is too lengthy and messy to post here.

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  • $\begingroup$ Can you link to a proof of the latter? $\endgroup$
    – davidlowryduda
    May 6, 2014 at 2:29
  • $\begingroup$ Is it posted as a preprint anywhere? On the arxiv or your advisor's site? $\endgroup$
    – davidlowryduda
    May 6, 2014 at 2:37
  • $\begingroup$ I conjecture that you can replace $6$ with any integer multiple of $6$ in the first formula, and statement is still true. But it seems kind of obvious. What do you do in this situation, is it better to state the technically more general result? $\endgroup$ May 6, 2014 at 5:58
  • $\begingroup$ I'd find it easier to get these Dedekind sum problems if you'd actually state them in terms of Dedekind sums. $\endgroup$ May 6, 2014 at 13:06
  • $\begingroup$ Is the last phrase the modern equivalent to Fermat's marginal note? /joking $\endgroup$
    – apnorton
    May 6, 2014 at 18:16

2 Answers 2

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Let $a,b \in \Bbb{R}$. Then each can be writ $a = a_i + a_f$, sim. for $b$, where $a_i = $ integer part, $a_f$ = fractional part. Then $ab = a_i b_i + a_i b_f + a_f b_i + a_f b_f$. Notice that if $(c \geq 0) \in \Bbb{Z}$ then $\{c + a\} = \{a\}$. Thus $\{ab\} = \{a_ib_f + a_f b_i + a_f b_f\}$. Notice that the fractional parts are $b_f = d/e$ with $d \lt e, d, e \in \Bbb{Z}$, sim. for $a$. Let's do some rewriting:

$\{a_ib_f + \dots + a_f b_f\} = \{a_i d/e + b_i f/g + (df)/(eg)\}$.

Okay, cancel that analysis but keep it just in casder e, since the numbers we are dealing with are $k \in \Bbb{Z}$ and a fraction $a/b$. So First assume $a(b-1) \lt b \equiv a = 1$. Then we have $\{a/b\} = a/b$ and we have $\{k a/b\} = \frac{\text{remainder of }ka/b}{b}$. But when you sum all the remainders over $k = 1 \dots b$, you get $$ \{a/b\} + \{2a/b\} + \dots + \{(b-1)a/b\}, \ \ \text{ since } \{ba/b\} = 0 $$

Now when you multiply a fraction by a constant $k \in \Bbb{Z}$ you get for instance $2 3/5 = 6/5 = 1 + 1/5$ or $k a/b = $ I give up! :)

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For $a = 1$ we have $\sum_{k =1}^b k\{ka/b\} = 1\{1/b\} + 2 \{2/b\} + \dots + (b-1)\{(b-1)/b\} = \sum_{k=1}^b k^2/b = 1/b \sum$(integers). And $\gcd(b,6)/\gcd(1,6) = \gcd(b,6)$ an integer. And the sum of squares formula from:

sum of sequence of squares article

has a $6$ on the bottom and a $b$ factor on the top, so I'm sure it has something to do with it!

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